Two 10 cm -long thin glass rods uniformly charged to +15nC are placed side by si

Question

Two 10 cm -long thin glass rods uniformly charged to +15nC are placed side by side, 4.0 cm apart. What are the electric field strengths E_1, E_2, and E_3 at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods? Specify the electric field strength E_2. Express your answer to two significant figures and include the appropriate units. E_2 = Specify the electric field strength E_3. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

The general expression for the electric field of a uniformly charged rod is,
E = K [1/x - 1/(x+L)]
K = 1/4o
= 8.98 x 109 Nm2C-2
lambda = Q/L = 15 x 10-9 / 0.1 = 15 x 10-8 C/m

a) E1
Electric field due to the rod in the left,
E1l = K [1/x - 1/(x+L)]
x = 0.01 m
E1l = (8.98 x 109) x (15 x 10-8) x [1/0.01 - 1/0.11] = 122454.55 N/C (towards right)
Electric field due to the rod in the right,
E1r = K [1/(0.04 - x) - 1/((0.004 -x) +L)]
= (8.98 x 109) x (15 x 10-8) x [1/0.03 - 1/0.13]
= 34538.46 (towards left)
The net electric field, E1 = E1l + E1r
= 122454.55 - 34538.46
= 87916.08 towards right.
= 88,000 N/C

2) E2
E2 = 0 since at the mid point between the rods (at x = 2), the electric field due to the left and the right rod become equal and they are oppositely directed.

3) E3
The magnitude of E3 is same as that of E1, but the direction is opposite.
E3 = - 88000 N/C (towards left.)

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